of 1 55
The Six-fold Nature of Reality
By
Ian Beardsley
Copyright © 2023 by Ian Beardsley
of 2 55
Contents
Abstract………………………………………………………………………….3
0.0 Six-Fold Symmetry and The Periodic Table…………………4
1.0 The Constant k………………………………………………………….9
2.0 Sixfold Nature of Protons…………………………………………..13
3.0 The Proton Radius…………………………………………………….19
4.0 Orbital Velocities as Six……………………………………………..21
5.0 Atomic Systems Mirror Planetary Systems………………….26
6.0 Rigorous Formulation of Proton-Seconds……………………34
7.0 Theory of Compounds……………………………………………….37
9.0 The Solar Magnetic Field…………………………………………..43
Conclusion…………………………………………………………………….50
Appendix 1…………………………………………………………………….52
Appendix 2……………………………………………………………………53
of 3 55
This work is the work A Six-fold Theory of Reality with an added section on the six-fold nature
of the solar magnetic field and its compatibility with life. We show that if we base reality on six-
fold symmetry we find it is described on every level for all of the building blocks from the
microcosmos to the macrocosmos from the atoms to the particles from which they are made
(protons) to the skeletal structure of life at it chemical basis the hydrocarbons, to the planets
and the moon. This is a rough sketch of the basis of reality as six-fold unfolding where we can
predict the radius of a proton and its charge. It all seems to waver around six, sometimes very
closely, but we should keep in mind there should be some room for play because for something
to have its function based on six it might only has to waver around it. We suggest from what we
find about the moon, that the curious fact that it perfectly eclipses the sun as seen from the
Earth, that it is there like that to have let us know since the time we were perhaps chipping
stones to make spearpoints around the fire at night, that we are not alone in the universe. We
formulate the idea of proton-seconds and find it suggests we can make a mathematical
formulation of the elements and their compounds aside from our chemical description.
of 4 55
0.0 Six-Fold Symmetry and The Periodic Table
We want to suggest that for a physical theory of reality the best basis for it is to begin with six-
fold symmetry. What do I mean by this? I like to start here, with a cube. It has six faces and as
such we see that six points describe a location in that putting six points at the center of each face
of the cube and connecting the points that oppose one another describes the location at its
center (O):
We let this represent
We notice each of these points on a face require two numbers in each face to describe them. So
twelve numbers become necessary:
We let this represent . This is a big reason that 6-fold
symmetry describes so much so well because 12 is the smallest
abundant number. It is evenly divisible by 1, 2, 3, 4, 6, and
1+2+3+4+6=16 is greater than 12.
Or, you can describe the point with three numbers but since there are six faces there are six
choices to do this:
We let this represent
6 ⋅ 1 = 6
2 ⋅ 6 = 12
3 ⋅ 6 = 18
of 5 55
From this we get our special ratios that are at the basis of the Natural systems which are
respectively in the the lengths of the unit lines in a plane and in space:
Which are irrational. We see them here:
These are:
But where is the golden ratio? It is in the regular pentagon:
Where is the unit 1? It is in the equilateral triangle, the most basic and stable structure, the first
2D object enclosed by straight lines:
1
2
+ 1
2
= 2
1
2
+ 2
2
= 3
2cos(π /4) = 2cos(45
∘
) = 2
2cos(π /6) = 2cos(60
∘
) = 3
2cos(π /5) = 2cos(36
∘
) =
5 + 1
2
2cos(π /3) = 2cos(30
∘
) = 1
of 6 55
of 7 55
Space is characterized by a 3 x 3 matrix because we can use it to do the cross product and form a
normal to two vectors in a plane. Let us pull out of the periodic table a 3 X 3 matrix from one we
will talk about in section 7.0 that we will see is pivotal to it.
A cube has six such faces, and each face finds it center (O) with two lines:
The center of a cube is found by 3 lines: one over, one in, one up:
3 × 3 = area = 9
2 × 3 = 6
of 8 55
The periodic table is periodic over 18 groups. 2, 3 are the
smallest primes, the lowest factors in which we can factor
numbers, we have in conclusion
2 ⋅ 3 = 6
3 ⋅ 3 = 9
2 ⋅ 9 = 18
3 ⋅ 6 = 18
of 9 55
1.0 The Constant k
We suggest there exists some k that serves as a constant that describes both the microcosmos
and macrocosmos from the proton, to the atoms, to planetary orbits. It is such that the square
root of it times the earth orbital velocity is 6, because we are guessing we are dealing with six-
fold symmetry as the basis of Nature. That is
Eq. 1.1
We have that k is
Eq. 1.2
This follows from what Warren Giordano noticed that
Eq. 1.3
Without the right units. I noticed since Avogadro’s number is that I
could introduce an equation of state for the periodic tables of the elements:
Eq. 1.4
Eq. 1.5
Let us say we were to consider Any Element say carbon . Then in general
We have
and
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have
12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus
And it follows that
k v
e
= 6
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
h(1 + α) ⋅ 10
23
= G
6.02 × 10
23
≈ 6 × 10
23
ℍ = 1
gram
atom
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
𝔼
ℂ
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6gram s
6proton s
N
A
=
6(6E 23proton s)
6gr a m s
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
of 10 55
We see in general since the atomic number Z is the number of protons in an atom that in general
this holds for all elements because
And,
Therefore we always have:
is a variable, the number of protons in multiplied by Avogadro’s number.
Put in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:
Equation. 1.6
While we have masses characteristic of the microcosmos like protons, and masses characteristic
of the macrocosmos, like the upper limit for a star to become a white dwarf after she novas (The
Chandrasekhar limit) which is 1.44 — More mass than that and she will collapse — we do not
have a characteristic mass of the intermediary world where we exist, a truck weighs several tons
and tennis ball maybe around a hundred grams. To find that mass let us take the geometric
mean between the mass of a proton and the mass of 1.44 solar masses. We could take the
average, or the harmonic mean, but the geometric mean is the squaring of the proportions, it is
the side of a square with the area equal to the area of the rectangle with these proportions as its
sides. We have:
Equation. 1.7
We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .
We have the intermediary mass is:
Equation 1.8
All we really need to do now is divide equation 1.6 by equation 1.7 and we get an even number
that is the six of our six-fold symmetry.
Equation. 1.9
The six of our six-fold symmetry.
We have something very interesting here. We have
𝔼
N
A
=
Z ⋅ 6E 23pr oton s
Z ⋅ gra m s
𝔼 =
Z ⋅ gra m s
Z ⋅ pr oton s
N
A
⋅ 𝔼 = 6E 23
N
A
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
M
⊙
= 1.98847E 30kg
m
p
= 1.67262E − 27kg
m
i
= (2.8634E 30)(1.67262E − 27) = 69.205kg
1
m
i
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 6.1092 ≈ 6
of 11 55
This is:
Equation 1.10.
Where k is a constant, given
Eq. 1.11 1
It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf
will not have its gravity overcome the electron degeneracy pressure and collapse. The non-
relativistic equation is:
Let us approximate 0.77 with 3/4. Since we have our constant
Equation. 1.12.
Equation 1.14.
Then
Equation 1.15.
Since our constant k in terms the Chandrasekhar limit is
Equation 1.16
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of
light squared, that is it represents the ground state. It is
Since
1
69.205kg
6kg
2
s
m
v
e
= 6
k v
e
= 6
k =
1
773.5
s
m
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
k =
1
m
2
i
h
1 + α
G
⋅ N
A
𝔼
m
i
= Mm
p
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
α
2
=
U
e
m
e
c
2
k v
e
= 6
of 12 55
We are suggesting the earth orbit is the ground state for our planetary system. We suggest it
holds for any planetary system because k as we will see is a natural constant that solves many
physical problems on many levels, not just planetary systems but atomic systems and the
particles that make them up.
of 13 55
2.0 Sixfold Nature of Protons
We now need to show that the fundamental particles that build reality are based on sixfold
symmetry for their mass, size, and charge and we want to do it in terms of gravity on the
macroscale, thus it has to use G the universal constant of gravitation and Planck’s constant h,
that quantizes energy on the microscale. I find I can do this as such:
The natural constants measure the properties of space and time. We can list some of them here:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Gravity is a property of space measured by the universal constant of gravity, G:
Equation 2.1
Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons
and neutrons) we will suggest given by:
Equation 2.2
Which describes mass per meter over time, which is:
Equation 2.3
It must be adjusted by the fine structure constant . It is my guess the factor should be which
is 18,769.:
Equation 2.4
Because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
Equation 2.5
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G: 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
G = 6.674E − 11N
m
2
s
2
= 6.674E − 11
m
3
s
2
⋅ kg
h
Gc
=
kg ⋅ s
m
1.82E − 16
kg ⋅ s
m
α
1
α
2
(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s /m
α
2
=
U
e
m
e
c
2
of 14 55
Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a
hypersphere. Thus we consider the surface area of a proton, :
Equation 2.6
We take the square root to get meters:
Equation 2.7
We multiply that with the value we have in equation 2.4:
Equation 2.8
We find that the mass of a proton realizes if we divide this by six seconds:
Equation 2.9
That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:
Equation 2.10
Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the
skeletons of organic matter, the so-called hydrocarbons:
Equation 2.11
Equation 2.12
If we divide 1E-26kgs by something greater than 6 seconds we get fractional protons. The rest of
the elements in the periodic table occur for dividing by something less that 1-second. It seems
the duration of a second is natural. If it is, since it was formed by a calendar based on
reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon
orbital mechanics. I find it is, that (See Appendix 1):
Equation 2.13
That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic
energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by
very small amounts over millions of years. The solar system has evolved towards this since the
explosion of life called the Cambrian, and will slowly decay away from it. But we need to derive
the second in terms of something else. For now we have the mass of a proton as:
S
p
S
p
= 4π r
2
p
= 8.72E − 30m
2
S
p
= 2.953E − 15m
(2.953E − 15m)(3.416E − 12
kg ⋅ s
m
) = 1.009E − 26kg ⋅ s
m
p
m
h
=
1E − 26
6secon d s
= 1.67E − 27kg
m
c
=
1E − 26kg ⋅ s
1secon d
= 1E − 26kg = 6pr oton s = 6m
p
m
p
=
1E − 26kg ⋅ s
6secon d s
m
c
=
1E − 26kg ⋅ s
1secon d
K E
moon
K E
earth
(Ear th Day) ≈ 1secon d
of 15 55
Equation 2.14
This way of looking at things is to say matter is that which has inertia. This means it resists
change in position with a force applied to it. The more of it, the more it resists a force. We
understand this from experience, but what is matter that it has inertia? In this analogy we are
suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As
such there has to be a normal vector holding the higher dimensional sphere in a lower
dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in
the plane there should be a force countering it proportional to the normal holding it in a lower
dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a
sphere. This counter force would be experienced as inertia. Since Planck’s constant h is a
measure of energy over time where space and time are concerned it must play a role. Of course
the radius of a proton plays a role since squared and multiplied by it is the surface area of our
proton embedded in space. The gravitational constant is force produced per kilogram over a
distance, thus it is a measure of how the surrounding space has an effect on the proton giving it
inertia. The speed of light c has to play a role because it is the velocity at which events are
separated through time. The mass of a proton has to play a role because it is a measurement of
inertia itself. And alas the fine structure constant describes the degree to which these factors
have an effect. We see the inertia then in equation 6 is six protons over 1 second, by dimensional
analysis.
m
p
=
3r
p
18α
2
4πh
Gc
4π
Fig. 1
of 16 55
Equation 2.15
That is 1 second gives carbon. We find six seconds gives 1 proton is hydrogen:
Equation 2.16
For time t greater than 6 seconds we have fractional protons. For t<6 we the have other
elements.
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number. We
make a program that looks for close to whole number solutions so we can create a table of values
for problem solving (Program in Appendix 2):
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. So we have the six-fold symmetry in the chemical
skeletons of life.
1
α
2
m
p
h 4πr
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4πr
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
1
α
2
m
p
h 4πr
2
p
Gc
of 17 55
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
Equation 2.17
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
x
ℝ
3
(x, y, z)
ℝ
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
−x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
− ct
2
)
i
··
x ∝ (ct
1
− ct
2
)
i
x
ℝ
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
− ct
2
)
i
g ⋅ m =
m
s
2
Fig. 2
of 18 55
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
Letting v= from our equation for k (page 12 is about k)
And we have
Equation 2.18
We get
Equation 2.19
g =
1
s
2
=
1
t
2
1
s
2
⋅
m
s
⋅ s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
⋅
q
2
x
2
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
⋅
1
k
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4πr
2
p
Gc
= q
2
h 4πr
2
p
Gc
=
(6.626E − 34)(4π)(0.833E − 15)
2
(6.674E − 11)(299,792, 459)
= 5.37E − 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E − 31) = 9.1435E − 19C
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71proton s ≈ 6pr otons
of 19 55
3.0 The Proton Radius
Thus we have the radius of a proton is given by carbon by evaluating at one second:
3.1.
But to get that we have to multiply by one second and we need one second in terms of the atom
for a theory of the proton. I find we can do that…
Where,…
, ,
Radius of hydrogen atom
Remember our constant k equation 1.16 (I find we have to divide by two somewhere and I think
this is because we are looking at packing protons here so they are offset by half their radius from
one another):
Equation 3.2. s
Since we have the equation of the radius of a proton is given by, by evaluating it at one second
which is carbon:
And 1 second in terms of the atom is given by
Then the equation for the radius of a proton is:
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
v =
r
t
t = 6s =
1
α
2
m
p
h 4πr
2
p
Gc
1
t
= α
2
m
p
Gc
h 4πr
2
p
v =
r
p
t
= α
2
m
p
Gc
h 4π
R
H
= 1.2E − 10m
t =
R
h
v
= R
h
⋅
1
α
2
m
p
h 4π
Gc
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12secon ds
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
p
⋅
R
H
N
A
𝔼
= 1.12secon ds
of 20 55
Equation 3.3
Let’s verify our equation:
Making the approximation 9/8~1 we can write equation 3.2 as (We suggest we have picked up
the fraction 9/8 by making several approximations):
Equation 3.4
Which gives
We form constants:
Equation 3.5
Equation 3.6
And we have the Equation:
Equation 3.7
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add
to this section the equation of the mass of a proton.
Equation 3.8
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
r
p
=
9
8
2
1
1.67E − 27
(6.626E − 34)(299,792, 459)
(6.674E − 11)4π
3
⋅
1.2E − 10
6.02E 23
= 0.93f m
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
= 8.26935E − 16m ≈ 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
ℍ
=
1
kg
(
kg
)
m
k =
hc
2π
3
G
= 6.93E − 9kg
R
H
N
A
𝔼
= 1.99E − 34m
r
p
m
p
= k
R
H
N
A
𝔼
m
p
=
1
r
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
of 21 55
4.0 Orbital Velocities as Six
We now find it convenient to describe the planetary orbital velocities as six in what we will call
natural units.
With the information we have for the earth-moon-sun orbital parameters we see we can make
the following equation in table form:
Consider minute times day:
Thus,..
The circle is divided into 360 units, each unit (each degree) is the distance the earth moves
around the Sun in a day, where a day is one turn of the earth on its its axis, and as such there are
360 such turns in the time it takes the earth to go around the sun approximately (365.25 days).
We have:
1 astronomical unit (AU) is the distance of the earth from the sun on the average, and is always
close to that because its orbit is approximately circular. We have
This is approximately the diameter of the Earth orbit. We define our variables:
Earth orbits:
Earth rotates:
Earth orbits:
Moon orbits:
Earth completes a 360 degree orbit yields:
orbit rotat ion orbit m oon
29.786k m minute 1d ay k ilom eter
secon d 27.83k m degree secon d
=
min ⋅ d ay
deg
⋅
k m
s
2
(min)(d a y) = 60(24 ⋅ 60 ⋅ 60) = 864,000sec
2
min ⋅ d ay
deg
⋅
k m
s
2
=
86,400s
2
deg
⋅
k m
s
2
= 86,400 ⋅
k m
deg
86,400 ⋅
k m
deg
⋅ 360
∘
= 311,040,000k m
311,040,000k m
149,598,000k m /AU
= 2.079AU
v
e
= 29.786
k m
s
ω
e
=
27.83k m
min
=
27.83k m
min
⋅
min
60sec
= 0.4638
k m
sec
θ
e
=
1d ay
deg
=
(24)(60)(60)
deg
= 14,400
sec
deg
v
m
=
1k m
sec
of 22 55
Where on the right it is in radians and is the radius of the Earth’s orbit. We have
This is (0.00618)360=2.225
0.00618 is
Where is the inverse of the golden ratio. .
We have:
Equation 4.1
Our base ten counting is defined
is defined
, such that
which is given by
Thus since the diameter of the Earth orbit is
Then its radius is
Equation 4.2
Since we have established a connection between the microcosmos and the macrocosmos we
would do well to introduce the units of AU (astronomical unit), year, solar masses. Thus we want
to know the universal gravitational constant in these units:
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅
360
∘
1AU
= 2π
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅ r
e
r
e
= 1AU
(29.786k m /s)(14,400s /deg)(1k m /s)
(0.4638k m /s)(149,598,000k m /AU )
360 = 2.225AU
ϕ
100
ϕ
ϕ = 0.618 =
1
Φ
ϕ
100
360
∘
= 2AU
10
0
= 1,10
1
= 1,10
2
= 100,...
ϕ
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
ϕ
100
360
∘
= 2AU
1
2
⋅
ϕ
100
360
∘
= 1AU
of 23 55
For a year we have
(365.25)(24)(60)(60)=31557600 seconds
And for an AU
1 AU=1.496E11m
We can immediately put this to work. In order for the earth to stay in orbit its centripetal force
must equal the gravitational force. Its orbital velocity must be given by:
Since in our theory we present R as:
Then
Equation 4.3
Which evaluates:
G = 6.67408E − 11
m
3
kg ⋅ s
2
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
G = 39.433
AU
3
M
⊙
⋅ year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
GM
⊙
= 40
AU
3
year
2
GM
⊙
4π
2
= 0.99885 ≈ 1
G
Mm
R
2
=
mv
2
r
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
v =
5
3
GM
ϕ
v =
5
3
(40)(1)
0.618
= 6
AU
year
of 24 55
Converting this to meters per second:
Which should be about right. The orbital velocity is given in the data tables as about 30,000m/s
is an average over a varying velocity due to the Earth’s slightly elliptical orbit.
Since we have
For the earth, we show now it is true for Venus as well. Its orbital distance VU (Venus units) is
1.082E11meters. It orbital period is 1.94E7s is the Venus year (VY).
=
Thus we have
Equation 4.4
We can then express all orbital velocities as 6, but to find their values in a formal system of
units, we need to convert from these natural units to something like kg/m/s. Thus logos
translated into a language we understand can best be done in a square array, or as a matrix
transformation. We have
Equation 4.5
Where is for the moon. You will find this gives , ,
All of which are correct within the variations of these velocities in their deviations from a
perfectly circular orbit.
6
AU
year
⋅
year
3.156E 7
⋅
1.496E11m
AU
= 28,441
m
s
1
2
⋅
ϕ
100
360
∘
= 1AU
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
V U
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
v
=
5
3
(40)(1)
ϕ
= 6
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
v
m
v
m
= 964.886m /s
v
e
= 28,443m /s
v
v
= 33,463.9m /s
of 25 55
Equation 4.6
Equations 4.7
Basically for the Earth
And in general for any orbiting body
Equation 4.8.
Where the right side is in the orbiting bodies’ natural units, which is its orbital radius taken as 1
divided by its year taken as 1. G and M are in these units as well, so G is constant and always 40,
and M is the mass orbited taken as 1. That is we could write
Equation 4.9.
And, again is the golden ratio conjugate is 0.618. . So,…
Equation 4.10.
Equation 4.11.
The orbits are about six because .
(
5
3
GM
ϕ
5
3
GM
ϕ
5
3
GM
ϕ
)
R
m
T
m
R
e
T
e
R
v
T
v
=
v
m
v
e
v
v
=
6
6
6
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1V U
v
e
=
5
3
GM
ϕ
= 6
AU
year
v
o
=
5
3
GM
ϕ
= 6
v
o
=
5
3
40
ϕ
= 6
ϕ
5/3 ≈ 3/4
v
o
=
3
4
GM
ϕ
= 6
v
o
=
3
4
40
ϕ
= 6
2π r = 2π (1) ≈ 6
of 26 55
5.0 Atomic Systems Mirror Planetary Systems
We have our equation for the Earth-Moon-Sun Orbital System
Equation 5.1
And our equation for the proton
Equation 5.2
Where is six protons is carbon (C). We also have that for the Earth orbital velocity
Equation 5.3
This results in
Equation 5.4
Which gives
Equation 5.5
This is
Because k is in s/m is 1/k=773.5 m/s. Thus we have something like a quantum state on the left
but for the planets equal to a quantum state for the proton on the right, and conclude k connects
the larger scale macrocosmos to the microcosmos of the atom. Or perhaps better to write:
Equation 5.6
Because then we have
K E
moon
K E
earth
(Ear th Day) ≈ 1secon d
1
α ⋅ 6m
p
h 4πr
2
p
Gc
= 1secon d
6m
p
k v
e
≈ 6
K E
moon
K E
earth
⋅ Ear th Day =
1
α
2
k v
e
m
p
h 4πr
2
p
Gc
k
(
v
e
K E
2
moon
K E
2
earth
Ear th Day
2
)
=
1
α
4
m
2
p
⋅
h 4πr
2
p
Gc
k (m ⋅ s) = s
2
k v
e
(
K E
2
moon
K E
2
earth
Ear th Day
2
)
=
1
α
4
m
2
p
⋅
h 4πr
2
p
Gc
s
2
= s
2
of 27 55
Because . It makes one think of how the energy of a wave is the amplitude
squared. Here amplitude then is in seconds and energy is the square of time.
Is actually equal to 1.2 seconds seconds squared, but we can say 1.2 seconds is a rough sketch for
the idea that it is one second, which you can do because in any physical theory there is room for
play for the physics to still serve its function. So we will evaluate it at one second. We have
On the right we have
This is an accuracy of about 95%. Equation 5.6 can be written
Equation 5.7.
Where 6 days is 6 rotations of the Earth. Just how accurate is this?
For the factor in terms of a proton we have
=
Since the mass of the earth and the mass of the moon do not vary, and the velocity of the moon
and the earth vary a little from aphelion to perihelion, their orbit are a little eccentric, we will
focus on the range of velocities.
k v
e
=
s
m
⋅
m
s
= 1
(
K E
2
moon
K E
2
earth
Ear th Day
2
)
k v
e
= 38.5
1
α
4
m
2
p
⋅
h 4πr
2
p
Gc
= 36.555
1
α
2
m
p
h 4πr
2
p
Gc
M
e
M
m
⋅
v
2
e
v
2
m
= 6d ays
1
α
2
m
p
h 4πr
2
p
Gc
=
18769
1.67262E − 27kg
(6.62607E − 34Js)(4π)(0.833E − 15m)
2
(6.67408E − 11Nm
2
/kg
2
)(299,792,459m /s)
(1.12213E 31)(5.37368E − 31) = 6.0299675 ≈ 6.03 ≈ 6.0secon d s
(6.03s)
M
e
M
m
⋅
v
2
e
v
2
m
= 6d ays
M
e
= 5.972E 24kg
M
m
= 7.34767E 22kg
6d ays = 6 (24)(60)(60) = 518400secon ds
M
e
M
m
= 81.2775
of 28 55
,
The measured values are, which are mean orbital velocities of the earth and moon:
,
Thus the accuracy of the equation is 29/32.5=90%. But, using the velocity of the moon at
aphelion which is 0.966 km/s and the velocity of the earth at perihelion which is 30.29km/s, we
have 30.29/0.966=31.356. This gives us an accuracy for our equation of 31.356/32.5=96.5%.
If in the planet that has life orbiting a star there is an indication to its intelligence that there is a
mystery before it, it is that for our star the Sun, the moon perfectly eclipses it as seen from the
Earth. This is because:
Which are approximately equal. As well we can look at it as:
Which are about the same as well. The interesting thing is that since our ratios are around
0.0025 and 0.0045, then…
I say this is interesting because this the ratio of the precious metal gold (Au) to that of silver (Ag)
by molar mass, these elements being used for religious and ceremonial jewelry, is the same 1.8:
(6.03s)(81.2775) = 490.103 ≈ 490secon ds
518400
490
= 1057.959 ≈ 1058
v
2
e
v
2
m
= 1058
v
e
v
m
= 32.5
v
e
= 29.78
k m
s
v
m
= 1.022
k m
s
29.78k m /s
1.022k m /s
= 29
(lu n ar − orbit)
(ear th − orbit)
=
384,400k m
149,592,870k m
= 0.00257
(lu n ar − r a diu s)
sol ar − ra diu s
=
1,738.1
696,00
= 0.0025
(lu n ar − r a diu s)
(lu n ar − orbit)
=
(1,738.1)
(384,400)
= 0.00452
sol ar − ra diu s
ear th − orbit
=
696,000
149,597,870
= 0.00465
0.0045
0.0025
=
9
5
= 1.8
Au
Ag
=
196.97
107.87
= 1.8
of 29 55
The sun is gold in color, the moon is silver in color.
Since the message from forces unknown is the moon, then we should take its size as a natural
unit and set it equal to one. If it has a radius of one, then the radius of the Sun is
The distance of the moon from the Earth is
We have
This is the that is the Solar radius to the lunar orbit around earth that is the molar mass of
gold to silver. This distance of the moon from the sun is about
Let us guess that the sun has to be the size that it is to be a yellow main sequence star. And that
the earth-moon system has to be 1 AU from the Sun to be in the habitable zone (to have water in
its liquid phase). Kepler’s Law of planetary motion is
Equation 5.8
For the Sun with T the orbital period of the planet and its distance from the Sun in
astronomical units, for circular orbits. For other stars we have to include a constant k involving
the masses of the bodies:
Equation 5.9
If the mass of the body orbiting the Star m is small compared to the mass of the star it is orbiting
we have
Equation 5.10
The Earth is the third planet, is in the habitable zone, and its distance from the Sun defines 1
AU. Thus we ask: What is k for other star systems? For stars on the main sequence their
luminosity is proportional to their masses raised to the power of 3.5 as an estimate. We have:
696,000k m
1,738k m
= 400
384,400k m
1,738k m
= 221.173 ≈ 220
400
220
= 1.818181...
9
5
149,592,870k m
1,738k m
= 8,608
T
2
= a
3
a
1
k
= G
M + m
4π
2
≈ G
M
4π
2
a
3
=
GM
4π
2
T
2
of 30 55
Equation 5.11
Where L is in solar luminosities and M is in solar masses:
Further if we say since the Earth is in the right zone to be habitable ( ) then if a star is 100
times brighter than the Sun by the inverse square law its habitable zone is is 10
times further from the star it orbits than the Earth is from the Sun. We have
Equation 5.12
Combining equations 2.4 and 2.5:
Equation 5.13
For another star system we can write equation 2.3
Equation 5.14.
Where n is the number of solar masses of the star and k has M in solar masses. Combining this
with equation 2.6 we have for the habitable zone of a star:
Equation 5.15
The luminosity of the sun is:
The separation between the earth and the sun is:
The solar luminosity at the earth is reduced by the inverse square law, so the solar constant is:
L =
(
M
M
⊙
)
3.5
L
⊙
= 3.9E 26J/s
M
⊙
= 1.98847E 30kg
ℍ
100 = 10AU
ℍ =
(
L
L
⊙
)
ℍ =
(
M
M
⊙
)
3.5
T
2
a
3
=
k
n
T
2
=
k
n
(
M
M
⊙
)
21
4
L
0
= 3.9 × 10
26
J/s
1.5 × 10
11
m
S
0
=
39 × 10
2
4π (1.5 × 10
11
)
= 1,370wat ts /m eter
2
of 31 55
That is the effective energy hitting the earth per second per square meter. This radiation is equal
to the temperature, to the fourth power by the steffan-boltzmann constant, sigma ( ), can
be called the temperature entering, the temperature entering the earth.
intercepts the earth disc, , and distributes itself over the entire earth surface, , while
30% is reflected back into space due to the earth’s albedo, a, which is equal to 0.3, so
But, just as the same amount of radiation that enters the system, leaves it, to have radiative
equilibrium, the atmosphere radiates back to the surface so that the radiation from the
atmosphere, plus the radiation entering the earth, is the radiation at the surface of
the earth, . However,
And we have…
a=0.3
So, for the temperature at the surface of the Earth:
T
e
σ
T
e
S
0
π r
2
4π r
2
σ T
e
4
=
S
0
4
(1 − a)
(1 − a)S
0
(
π r
2
4π r
2
)
σ T
a
4
σ T
e
4
σ T
s
4
σ T
a
4
= σ T
e
4
σ T
s
4
= σ T
a
4
+ σ T
e
4
= 2σ T
e
4
T
s
= 2
1
4
T
e
σ T
e
4
=
S
0
4
(1 − a)
σ = 5.67 × 10
−8
S
0
= 1,370
1,370
4
(0.7) = 239.75
T
e
4
=
239.75
5.67 × 10
−8
= 4.228 × 10
9
T
e
= 255Kelvin
T
s
= 2
1
4
T
e
= 1.189(255) = 303Kelvin
of 32 55
Let’s convert that to degrees centigrade:
Degrees Centigrade = 303 - 273 = 30 degrees centigrade
And, let’s convert that to Fahrenheit:
Degrees Fahrenheit = 30(9/5)+32=86 Degrees Fahrenheit
In reality this is warmer than the average annual temperature at the surface of the earth, but in
this model, we only considered radiative heat transfer and not convective heat transfer. In other
words, there is cooling due to vaporization of water (the formation of clouds) and due to the
condensation of water vapor into rain droplets (precipitation or the formation of rain).
The point we want to make is that the Earth has to be the distance it is from the Sun, and the
Sun has to be the size that it is. We have that
Which allows the moon to perfectly eclipse the Sun. However, this means
If the solar radius and earth orbit had to have been what they are for the Earth to be habitable,
the lunar radius and solar radius can be anything’s to satisfy this equation. This is the part
where the messenger, who we hypothesize put the moon in the sky such that it would eclipse the
sun, had a criterion to choose the size of the lunar radius and the size of its orbit. How, under
this hypothesis did it or they determine their proportions? I think the best answer would be in
equation 5.6:
Which would mean it was because of the size of the proton ( ) and its mass ( ). We derived
these values and the charge of a proton from the properties of space and time.
We have to explain the meaning of the earth day in the equation, we might be able to do this by
breaking the whole equation down into parts.
If is the angular velocity of the earth, is the mass of the moon, is the mass of the earth,
is the orbital velocity of the moon, is the orbital velocity of the earth, then we have
There are 86,400 seconds in an earth day so, the angular velocity of the earth is
SolarRa diu s
Lun arRa diu s
= 400 =
Ear thOrbit
Lun arOrbit
= 389
400 ≈ 389
Lun arRa diu s
Lun arOrbit
=
SolarRa diu s
Ear thOrbit
k v
e
(
K E
2
moon
K E
2
earth
Ear th Day
2
)
=
1
α
4
m
2
p
⋅
h 4πr
2
p
Gc
r
p
m
p
ω
e
M
m
M
e
v
m
v
e
k v
e
⋅
M
2
m
M
2
e
⋅
v
4
m
v
4
e
(
2π
ω
e
)
2
=
1
α
4
m
2
p
⋅
h 4πr
2
p
Gc
of 33 55
Since we have
There are 31,557,600 seconds in a year, so our value above is:
Since there are 365.25 days in a year that is
It is 6 days, which is six rotations of the earth. The whole reason then, for the hypothetical
messenger’s equation, could be to maintain our six-fold symmetry.
ω
e
=
2π
86,400s
= 7.2722E − 5
ra d
s
ω
2
e
= 5.2885E − 9
ra d
2
s
2
k v
e
= 38.5
(38.5)(4π
2
ra d
2
) = 1,520ra d
2
M
2
m
M
2
e
⋅
v
4
m
v
4
e
(
1,520ra d
2
ω
2
e
)
=
M
2
m
M
2
e
⋅
v
4
m
v
4
e
(2.874E11s
2
)
M
m
M
e
⋅
v
2
m
v
2
e
(536,097s) =
1
α
2
m
p
h 4πr
2
p
Gc
536,097s
31,557,600s /yr
= 0.017years
(0.017)(365.25d ays) = 6d ays
of 34 55
6.0 Rigorous Formulation of Proton-Seconds
We can actually formulate this differently than we have. We had
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2 are
lower and upper limits in an integral, then we have:
Equation 6.1
This Equation is the generalized equation we can use for solving problems. Essentially we can
rigorously formulate the notion of proton-seconds by considering
Equation 6.2
Is protons-seconds squared where current density is and ( can also be
). We say
Equation 6.3
Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is
Equation 6.4
Dividing Equation 6.2 through by t:
Equation 6.5
Which is proton-seconds. Dividing through by t again:
1
t
1
1
α
2
m
p
h 4πr
2
p
Gc
= 6proton s
1
t
6
1
α
2
m
p
h 4πr
2
p
Gc
= 1proton
1
α
2
m
p
h 4πr
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
∫
t
qdt = t
2
∬
S
ρ(x, y, z)d x d y
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρdV
ℕ =
1
α
2
m
p
h 4πr
2
p
Gc
1
α
2
m
p
h 4πr
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z)d x d y
of 35 55
Equation 6.6
We see that if where and then J is I/m2 (current per square meter)
is analogous to amperes per per square meter which are coulombs per second through a surface.
Thus we are looking at a number of protons per second through a surface. Thus we write:
Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:
Equation 6.7
So as an example,…
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter.
!
1
α
2
m
p
h 4πr
2
p
Gc
∫
t
dt
t
2
= proton s
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h 4πr
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
1
α
2
m
p
h 4πr
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
1
α
2
m
p
h 4πr
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
proton s
secon d
J(x, y, z) = (0,0, J ) = − J
k
d
S = d x d y
k
J ⋅ d
S = (0,0, J ) ⋅ (0,0, d x d y) = − Jd x d y
of 36 55
We are now equipt to do computations in proton-seconds. We use equation 3.6 from two to
three, the smallest prime numbers that multiply to make six-fold symmetry in our hexagonal
proton that we found described its radius (My feeling is we introduce the factor of 2 because
carbon is 6 protons +6 neutrons and 2 times 6 is twelve):
Now we integrate from phosphorus is 15 protons=0.396 seconds to aluminum is 13 protons =
0.462 seconds which is to integrate across silicon (divide your answer 10 by 2 to get protons):
By what value would you like to increment?: 0.006
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
251.2490 protons 0.024000 seconds 0.248978 decpart
59.1174 protons 0.102000 seconds 0.117416 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
40.1998 protons 0.150000 seconds 0.199844 decpart
37.2221 protons 0.162000 seconds 0.222080 decpart
27.1621 protons 0.222000 seconds 0.162058 decpart
25.1249 protons 0.240000 seconds 0.124907 decpart
20.0999 protons 0.300000 seconds 0.099916 decpart
17.0338 protons 0.354000 seconds 0.033823 decpart
16.2096 protons 0.372000 seconds 0.209604 decpart
15.2272 protons 0.396000 seconds 0.227202 decpart
14.1549 protons 0.426000 seconds 0.154862 decpart
13.2236 protons 0.456000 seconds 0.223620 decpart
13.0519 protons 0.462000 seconds 0.051885 decpart
12.1084 protons 0.498000 seconds 0.108375 decpart
11.1666 protons 0.540000 seconds 0.166615 decpart
11.0439 protons 0.546000 seconds 0.043906 decpart
10.1515 protons 0.594000 seconds 0.151471 decpart
Equation 6.8
Silicon can be doped with phosphorus to make negative (n-type) silicon that semi-conducts thus
enabling the construction of logic circuits that you can use to make computing machines. But
this must be joined with positive (p-type) silicon which usually uses boron, but boron is in the
same group as aluminum, just above it. This results in a theory for AI elements as mathematical
constructs, that we will go into now.
2
α
2
m
p
h 4πr
2
p
Gc
∫
t
dt
t
2
= proton s
2
α
2
m
p
h 4πr
2
p
Gc
∫
t
Al
t
P
dt
t
3
= − 6(6.376 − 4.685) = 5proton s /secon d = boron
of 37 55
7.0 Theory of Compounds
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the
elements. As you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and
germanium are in group 14 meaning they have 4 valence electrons and want 4 for more to attain
noble gas electron configuration. If we dope Si with B from group 13 it gets three of the four
electrons and thus has a deficiency becoming positive type silicon and thus conducts. If we dope
the Si with P from group 15 it has an extra electron and thus conducts as well. If we join the two
types of silicon we have a semiconductor for making diodes and transistors from which we can
make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the
right of Si but doping agent B is not directly to the left, aluminum Al is. This becomes important.
I call (As-Ga) the differential across Ge, and (P-Al) the differential across Si and call Al a dummy
in the differential because boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with
subscripts that tell what element it is and it properties, I have done this:
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the
first row and has 4 valence electrons making it carbon (C). I believe that the AI elements can be
organized in a 3 by 3 matrix makes them pivotal to structure in the Universe because we live in
three dimensional space so the mechanics of the realm we experience are described by such a
matrix, for example the cross product. Hence this paper where I show AI and biological life are
mathematical constructs and described in terms of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are
carbon and nitrogen respectively, there is every reason to proceed with this paper if the idea is to
show not only are the AI elements and biological elements mathematical constructs, they are
described in terms of one another. We see this because the first row is ( B, C, N) and these
happen to be the only elements that are not core AI elements in the matrix, except boron (B)
which is out of place, and aluminum (Al) as we will see if a dummy representative makes for a
mathematical construct, the harmonic mean. Which means we have proved our case because the
first row if we take the cross product between the second and third rows are, its respective unit
vectors for the components meaning they describe them.
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
of 38 55
The Computation
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological
elements C and N being the unit vectors are multiplied by the AI elements, meaning they
describe them. But we have to ask; Why does the first row have boron in it which is not a core
biological element, but is a core AI element? The answer is that boron is the one AI element that
is out of place, that is, aluminum is in its place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon
(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic
(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)
and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they
need doping agents from group 13 (three valence electrons) like boron and gallium, and to have
negative type silicon and germanium they need doping agents from group 15 like phosphorus
and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a
different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al)
is in boron’s place. This results in an interesting equation.
Equation 7.1
A = (Al, Si, P)
B = (G a, G e, A s)
A ×
B =
B
C
N
Al Si P
Ga Ge As
= (Si ⋅ A s − P ⋅ Ge)
B + (P ⋅ G a − Al ⋅ As)
C + (Al ⋅ Ge − Si ⋅ Ga)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g/m ol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g/m ol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsinθ = (50)(126)sin8
∘
= 877.79
877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol
Si(As − G a) + Ge(P − Al )
SiG e
=
2B
Ge + Si
of 39 55
The differential across germanium crossed with silicon plus the differential across silicon
crossed with germanium normalized by the product between silicon and germanium is equal to
the boron divided by the average between the germanium and the silicon. The equation has
nearly 100% accuracy (note: using an older value for Ge here, it is now 72.64 but that makes the
equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic
mean between the semiconductor elements (by molar mass):
Equation 7.2
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
We can make this into two integrals:
Equation 7.3
Equation 7.4
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(A s − G a) +
Ge
B
(P − Al ) =
2SiGe
Si + Ge
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d xdy ≈
1
Ge − Si
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Si
B
(As − Ga)d yd z ≈
1
3
1
(Ge − Si)
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Ge
B
(P − Al )d x dz ≈
2
3
1
(Ge − Si)
∫
Ge
Si
yd y
of 40 55
If in the equation (The accurate harmonic mean form):
Equation 7.5
We make the approximation
Equation 7.6
Then the Stokes form of the equation becomes
Equation 7.7
Thus we see for this approximation there are two integrals as well:
Equation 7.8
Equation 7.9
By making the approximation
In
We have
Equation 7.10
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
dydz =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − Ga)d yd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )d yd z =
2
3
∫
Ge
Si
dz
2SiGe
Si + Ge
≈ Ge − Si
Si(As − Ga)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ Ge
ΔSi
ΔS
= B
of 41 55
is the differential across Si, is the differential across Ge
and is the vertical differential.
We say (Phi) is given by
and
And
=1.618
=0.618
(phi) the golden ratio conjugate. We also find
Equation 7.11
We can go straight down group 14 to form the rest of our differentials:
It is amazing how accurately we can fit these differentials with an exponential equation
for the upward increase. The equation is
This is the halfwave:
Since B/Al=10.81/26.98=
0.40
And Ag/Cu =107.87/63.55=
1.697
ΔSi = P − Al
ΔGe = As − Ga
ΔS = Ge − Si
Φ
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b/a
ϕ
(ϕ)ΔGe + (Φ)ΔSi = B
ΔC = N − B = 14.01 − 10.81 = 3.2
ΔSi = P − Al = 30.97 − 26.98 = 3.99
ΔGe = As − Ga = 74.92 − 69.72 = 5.2
ΔSn = Sb − In = 121.75 − 114.82 = 6.93
ΔPb = Bi − Tl = 208.98 − 204.38 = 4.6
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
of 42 55
Where B and Al are the
differential across silicon
and Ag and Au are the finest
conductors of electricity, we
have
Equation 7.12
y(x) = e
(B/Al)x
+
Ag
Cu
of 43 55
8.0 The Solar Magnetic Field
We model the formation of the solar system from a slowly rotating gas cloud, a nebula of
gaseous molecules, that collapses into a flat disc with a protostar at its center. The star turns on
and blows lighter elements far away, like hydrogen and helium, from which form the gas giants,
like Jupiter and Saturn, and the heavier elements stay closer in, like iron and silicates, from
which form the terrestrial planets like Venus, Earth, and Mars form. There are basically three
factors that determine its structure, the inward gravity, the pressure gradient outward which
balances with the inward gravity, and the outward inertial forces from the planets’ orbits. The
flattened rotating disc is broken up into rings each that has a mass spread out over it from which
the planets form. We estimate the ring associated with the Earth, had in its lower limit 230 earth
masses spread over it for the Earth to form. We further estimate that the Venus ring had a mass
spread over it of 230 Venus masses for Venus to form, and the Mars ring similarly had 230 Mars
masses spread out over it for Mars to form. The asteroid belt had about 200 of it masses, and the
Jovian planets 5, 8, 15, and 20 masses of each respectively. For Mercury it requires a factor of
about 350 because it is mostly iron condensations with incomplete silicon condensations.
Plotting these logarithmically we get the exponent of r, the distance of a planet from the sun is
-1.5 so that the density distribution of the protoplanetary disc is:
Giving a mass
With pressure gradients playing the key role in the formation of solar system, less attention is
payed to the magnetic field of the Sun. However, in the older literature, one of the pioneer’s of
this aspect found something very interesting concerning it. He was Alfven (1942). At the time
people were suggesting instead of the solar system forming from a rotating nebula, rather the
sun came into existence not at the same time at the center of the disc, but rather passed through
clouds and captured material after already existing. He figured for the captured material its
inward component v, and density , at a distance r from the sun, had to conserve mass, which
required:
He figured as the velocities of the atoms got closer to the sun, were moving then faster, collisions
would increase, and so temperature would go up, ionizing the atoms and therefore ionized, the
magnetic field becomes important. He considered for simplicity the solar magnetic field was
generated by a dipole moment , a vector quantity, and that a particle moving in the plane of
that vector with mass m and charge q, would have all of both the gravitational and magnetic
forces in that plane, so the problem becomes two-dimensional and required only the and , of
polar coordinates. The differential equations of its motion would be:
σ (r) = σ
0
r
−3/2
σ
0
= 3300
M =
∫
2π
0
∫
r
h
r
s
σ (r)r dr dθ
ρ
dM
dt
= 4π r
2
ρv
μ
θ
r
of 44 55
Equation 8.1.
And,
Equation 8.2.
We can integrate equation 8.2 with the boundary condition that the angular momentum of the
particle is zero at large distances from the sun to get:
Equation 8.3.
And substitute it into equation equation 8.1 for to get
Equation 8.4.
Which we can write
Equation 8.5.
We then integrate this with respect to r with the boundary condition that at large r and get
Equation 8.6.
He then notices there is another value for which . It is
Equation 8.7.
This is interesting because it means the particle can never approach the Sun closer than this
value, and it depends only on the value q/m, the charge to mass ratio of a particle. He took this
as hydrogen because ionized it is a proton, for which q/m is well defined. He estimated what the
magnetic field of the Sun could have been in this earlier stage of its life, and adjusted for the fact
that hydrogen doesn’t ionize until it reaches a velocity of 5E4 m/s and found that was the
region occupied by the major planets which are Jupiter and Saturn mostly made of hydrogen
and helium.
Certainly today we don’t see the planets as having formed from material gathered by the Sun in
its journey, but rather think the Sun and planets formed at the same time from a cloud that
collapsed into a rotating flat disc. And indeed, there may be stars in the galaxy that pass through
clouds and gather material, and indeed Alfven’s equations would hold preventing ionizing
clouds from falling into their star. But we can also apply his equation to our Sun today, for which
we know a great deal about its magnetic field, which also happens to be an important thing to
study and for which we have satellites in the Lagrange points, where the Earth’s gravity cancels
m
··
r = −
GM
⊙
m
r
2
+
qμ
·
θ
r
2
+ mr
·
θ
2
m
r
d(r
2
·
θ )
dt
= −
qμ
·
r
r
3
mr
2
·
θ = −
∫
qu dr
r
2
=
qu
r
·
θ
m
··
r = −
GM
⊙
r
2
+
2q
2
μ
2
m
2
r
5
·
r
d
·
r
dr
= −
GM
⊙
r
2
+
2q
2
μ
2
m
2
r
5
·
r = 0
·
r
2
=
2GM
⊙
r
−
q
2
μ
2
m
2
r
4
·
r = 0
r
c
=
(
q
2
μ
2
2GM
⊙
m
2
)
1/3
r
c
of 45 55
with that of the Sun, where the orbits are very stable, so we can understand the solar magnetic
field. It is a complex field, that interacts with the Earth’s magnetosphere, and we need to predict
solar maximums, so we have warning as to whether there will be a magnetic storm that will
knock out our electrical grid and internet, ahead of time.
During solar minimum the solar magnetic field has closed lines, that flow out one pole and into
the other. The dipole field of the sun is about 50 Gauss. There are 10,000 Gauss in a Tesla, so
that is 5E-8 Tesla. That is the magnetic field strength where the field goes into the poles. The
total magnetic field of the Sun at the Earth, is all the components taken together, which are
, , and . The important component is , because it runs north-south, so it is
perpendicular to the ecliptic, the path traced out by the sun due to the earth’s orbit. It is the
component that interacts with the Earth magnetosphere, and when it points southward, it will
connect with the Earth’s magnetosphere which points northward so the solar poles flow into the
Earth poles and the Earth field then gets disrupted allowing particles from the solar wind to rain
down along Earth magnetic field lines causing the Aurora. The solar magnetic field doesn’t
always stay around the Sun itself, but the solar wind carries it through the solar system until it
collides with the interstellar medium reaching the heliopause. Thus the Sun creates the
Interplanetary Magnetic File (IMF) which has a spiral shape because the Sun rotates once about
every 25 days. But the upshot is that at Earth we have
Moderate Magnetic Field: 10 nT
Strong Magnetic Field: 20 nT
Very Strong Magnetic Field: 30 nT
For our purposes we want to return to equation 7:
B
t
B
x
B
y
B
z
B
z
of 46 55
And ask just what is , because in the time that Alfven was working we worked with magnetic
fields differently, aside from his equation uses a trick, which we still use today, and that is to
consider the magnetic field a dipole. To consider it like this is to say there are two monopoles
opposite in polarity. According to Maxwell’s equation we cannot have magnetic monopoles,
though they are predicted by some modern theories, they have never been found. The trick is in
that by treating the North magnetic pole and South magnetic pole as separate magnetic charges
is to treat them like we do electric charges, the charge of a proton and the charge of an electron,
which can be convenient for making computations, but don’t exist that we know of. So we will
solve equation 7 for , and see what its units are so we can understand what it represents and we
will let m be the mass of a proton and q the charge of a proton. We get:
Equation 8.8.
We can write these units as, by taking Coulombs (C) equal to
Equation 8.9.
This is units of force per current density, which makes sense because a flowing current creates a
force. We can also write it:
Equation 8.10.
Which is energy per magnetic field strength in that the SI units of magnetic field strength is
amps per meter. This tells us:
Equation 8.11.
Thus we will use the energy as ionization of hydrogen, the energy to remove its electron and
make it a proton:
We have
, , ,
r
c
=
(
q
2
μ
2
2GM
⊙
m
2
)
1/3
μ
μ
r
3
GM
⊙
m
2
p
q
2
p
= μ =
m
3
⋅ kg
C ⋅ s
a m ps ⋅ secon d s
2
kg
m
s
2
⋅
m
2
a m ps
kg
m
2
s
2
⋅
m
a m ps
μ =
Energ y
Magnet icFiel dSt r ength
H ⟶ H
+
+ e
−
= 1proton = 2.18E − 18J
q
p
= 1.6E − 19C
m
p
= 1.67E − 27kg
G = 6.67408E − 11N
m
2
kg
2
M
⊙
= 1.989E 30kg
of 47 55
We want to look at the Sun as having a current flowing around its equator in a loop with its
radius
We find for the dipole field of the Sun at 50 Gauss=5E-3T, which is about 100 times stronger
than the Earth magnetic field, that this is a current I=5.5362E12 amperes driving the solar
magnetic dipole. This gives us that since the Earth orbit (1AU=1.495979m):
Equation 8.12.
Equation 8.13.
Or,…
Equation 8.14.
The radius of a proton is 0.833E-15m. We have that r is:
Equation 8.15.
R
⊙
= 6.957E8m
I
1AU
=
5.5362E12a mps
1.496E11m
= 37.0 a mps /m
μ =
Ioni zation Energ y
Magnet icFiel dSt r ength
=
2.18E − 19J
37.0A m ps /m
= 5.892E − 20
J ⋅ A
m
r =
(
q
2
p
μ
2
2GM
⊙
m
2
p
)
1/3
= 4.9324E − 15m
r
r
p
= 5.92 ≈ 6Proton Ra dii = carbon
of 48 55
Our six-fold symmetry unfolding. This is again the carbon the core element of life. We see it
provided for by the Sun’s magnetic field.
of 49 55
Let us write the computation as one equation, and verify it. We have
Where
=Ionization energy of Hydrogen
=3.47E-39 (correct)
Equation 8.16
(0.000034574)(3.47E-39)=1.19972E-43
1.19972E-43^(1/3)
=4.932E-15
So as you can see equation 8.16 is correct.
r =
(
q
2
p
μ
2
2GM
⊙
m
2
p
)
1/3
= 4.9324E − 15m
I
⊙
=
2B
⊙
R
⊙
μ
0
I
1AU
=
5.5362E12a mps
1.496E11m
= 37.0am ps /m
1AU = r
e
μ =
Ioni zation Energ y
Magnet icFiel dSt r ength
=
2.18E − 18J
37.0A m ps /m
= 5.892E − 20
J ⋅ A
m
IE
H
I
⊙
1AU
=
2B
⊙
R
⊙
μ
0
1
r
e
μ
2
= (5.982E − 20)
2
= 3.47E − 39
μ
2
=
(IE
H
)
2
μ
2
0
r
2
e
4B
2
⊙
R
2
⊙
=
(2.18E − 18)
2
(12.56637E − 7)
2
(1.496E11)
2
4(5E − 3)
2
(6.957E8)
2
6r
p
=
(
q
2
p
2GM
⊙
m
2
p
⋅
(IE
H
)
2
μ
2
0
r
2
e
4B
2
⊙
R
2
⊙
)
1/3
6r
p
=
(1.6E − 19C )
2
2(6.67408E − 11N
m
2
kg
2
)(1.989E 30kg)(1.67E − 27)
2
⋅
(2.18E − 18J )
2
(12.56637E − 7T ⋅ m /A)
2
(1.496E11m)
2
4(5E − 3T )
2
(6.957E8m)
2
1/3
of 50 55
Conclusion
Indeed we have shown that we can predict the radius of a proton and its mass in terms of the
properties of space:
Which gives
And the interesting thing is they do not yet have an equation for the radius of a proton, yet we
have it here and it doesn’t just use Planck’s constant, h, of the indeterministic microcosmos, but
uses the gravitational constant, G, of the deterministic macrocosmos. We have further shown
our theory of inertia is based on six-fold symmetry is centered around hydrocarbons, the
skeletons of life chemistry
We find 1 second gives
We find six seconds gives 1 proton is hydrogen:
We further find this same sixfold symmetry describes not just proton systems (atoms) but our
solar system
We also have that
Where,
And
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
= 8.26935E − 16m ≈ 0.827f m
1
α
2
m
p
h 4πr
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4πr
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
(
6 6 6
)
r
m
/T
m
r
e
/T
e
r
v
/T
v
=
v
m
v
e
v
v
k v
e
= 6
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
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We further find that for the moon
Where six days is six rotations of the Earth. This shows the earth/moon/sun system and the
proton are structured around one another in terms of six-fold symmetry. Since there are 360
degrees in a circle, and the earth year is close to 360 rotations of the earth (365.25) the earth
orbit can be divided into six sixty degree sectors, the earth moving through about one degree a
day in its orbit.
We have further found that the solar magnetic field follow through with the same six-fold
symmetry, accommodating carbon:
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4πr
2
p
Gc
= q
2
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71proton s ≈ 6pr otons
1
α
2
m
p
h 4πr
2
p
Gc
M
e
M
m
⋅
v
2
e
v
2
m
= 6d ays
6r
p
=
(
q
2
p
2GM
⊙
m
2
p
⋅
(IE
H
)
2
μ
2
0
r
2
e
4B
2
⊙
R
2
⊙
)
1/3
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Appendix 1
We suggested the second was a natural unit, and that, if it was, should be in the orbital
mechanics of the earth because the second comes from the calendar, which is based on the
orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the
rotation of the Earth. We found that it was as the following:
Let’s show that…
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.
K E
moon
K E
earth
(Ear th Day) ≈ 1secon d
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
2π (3.85E 8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
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Appendix 2
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number. We
make a program that looks for close to whole number solutions so we can create a table of values
for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. Here is the code for the program:
1
α
2
m
p
h 4πr
2
p
Gc
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#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
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The Author
